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Numbers formed by flipping common set bits in two given integers

  • Difficulty Level : Medium
  • Last Updated : 08 Apr, 2021

Given two positive integers A and B, the task is to flip the common set bitsin A and B.

Examples:

Input: A = 7, B = 4 
Output: 3 0 
Explanation: 
The binary representation of 7 is 111 
The binary representation of 4 is 100 
Since the 3rd bit of both A and B is a set bit. Therefore, flipping the 3rd bit of A and B modifies A = 3 and B = 0 
Therefore, the required output is 3 0

Input: A = 10, B = 20 
Output: 10 20

Naive Approach: The simplest approach to solve this problem is to check if the ith bit of both A and B is a set or not. If found to be true, then flip the ith bit of both A and B. Finally, print the updated values of both A and B.



Below is the implementation of the above approach:

C++




// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to flip bits of A and B
// which are set bits in A and B
void flipBitsOfAandB(int A, int B)
{
    // Iterater all possible bits
    // of A and B
    for (int i = 0; i < 32; i++) {
 
        // If ith bit is set in
        // both A and B
        if ((A & (1 << i)) && (B & (1 << i))) {
 
            // Clear i-th bit of A
            A = A ^ (1 << i);
 
            // Clear i-th bit of B
            B = B ^ (1 << i);
        }
    }
 
    // Print A and B
    cout << A << " " << B;
}
 
// Driver Code
int main()
{
    int A = 7, B = 4;
 
    flipBitsOfAandB(A, B);
 
    return 0;
}

Java




// Java program to implement
// the above approach
import java.util.*;
   
class GFG{
   
// Function to flip bits of A and B
// which are set bits in A and B
static void flipBitsOfAandB(int A, int B)
{
     
    // Iterater all possible bits
    // of A and B
    for(int i = 0; i < 32; i++)
    {
         
        // If ith bit is set in
        // both A and B
        if (((A & (1 << i)) &
             (B & (1 << i))) != 0)
        {
             
            // Clear i-th bit of A
            A = A ^ (1 << i);
  
            // Clear i-th bit of B
            B = B ^ (1 << i);
        }
    }
  
    // Print A and B
    System.out.print(A + " " + B);
}
   
// Driver Code
public static void main(String[] args)
{
    int A = 7, B = 4;
  
    flipBitsOfAandB(A, B);
}
}
 
// This code is contributed by code_hunt

Python3




# Python3 program to implement
# the above approach
 
# Function to flip bits of A and B
# which are set in both of them
def flipBitsOfAandB(A, B):
     
    # Iterate all possible bits of
    # A and B
    for i in range(0, 32):
         
        # If ith bit is set in
        # both A and B
        if ((A & (1 << i)) and
            (B & (1 << i))):
             
            # Clear i-th bit of A
            A = A ^ (1 << i)
 
            # Clear i-th bit of B
            B = B ^ (1 << i)
             
    print(A, B)
 
# Driver Code 
if __name__ == "__main__" :
     
    A = 7
    B = 4
     
    flipBitsOfAandB(A, B)
     
# This code is contributed by Virusbuddah_

C#




// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to flip bits of A and B
// which are set bits in A and B
static void flipBitsOfAandB(int A, int B)
{
     
    // Iterater all possible bits
    // of A and B
    for(int i = 0; i < 32; i++)
    {
         
        // If ith bit is set in
        // both A and B
        if (((A & (1 << i)) &
             (B & (1 << i))) != 0)
        {
             
            // Clear i-th bit of A
            A = A ^ (1 << i);
 
            // Clear i-th bit of B
            B = B ^ (1 << i);
        }
    }
 
    // Print A and B
    Console.Write(A + " " + B);
}
 
// Driver Code
public static void Main(string[] args)
{
    int A = 7, B = 4;
 
    flipBitsOfAandB(A, B);
}
}
 
// This code is contributed by chitranayal

Javascript




<script>
// javascript program to implement
// the above approach
 
 
// Function to flip bits of A and B
// which are set bits in A and B
function flipBitsOfAandB(A , B)
{
     
    // Iterater all possible bits
    // of A and B
    for(i = 0; i < 32; i++)
    {
         
        // If ith bit is set in
        // both A and B
        if (((A & (1 << i)) &
             (B & (1 << i))) != 0)
        {
             
            // Clear i-th bit of A
            A = A ^ (1 << i);
  
            // Clear i-th bit of B
            B = B ^ (1 << i);
        }
    }
  
    // Print A and B
    document.write(A + " " + B);
}
   
// Driver Code
var A = 7, B = 4;
 
flipBitsOfAandB(A, B);
 
// This code is contributed by Princi Singh
 
</script>
Output: 
3 0

 

Time Complexity: O(32) 
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is based on the following observations:

Find the bits which are set bits in both A and B using (A & B). 
Clear the bits of A which are set bits in both A and B using A = (A ^ (A & B)) 
Cleat the bits of B which are set bits in both A and B using B = (B ^ (A & B)) 
 

Follow the steps below to solve the problem:

  • Update A = (A ^ (A & B)).
  • Update B = (B ^ (A & B)).
  • Finally, print the updated values of A and B.

Below is the implementation of the above approach:

C++




// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to flip bits of A and B
// which are set in both of them
void flipBitsOfAandB(int A, int B)
{
 
    // Clear the bits of A which
    // are set in both A and B
    A = A ^ (A & B);
 
    // Clear the bits of B which
    // are set in both A and B
    B = B ^ (A & B);
 
    // Print updated A and B
    cout << A << " " << B;
}
 
// Driver Code
int main()
{
    int A = 10, B = 20;
 
    flipBitsOfAandB(A, B);
 
    return 0;
}

Java




// Java program to implement
// the above approach
import java.util.*;
    
class GFG{
    
// Function to flip bits of A and B
// which are set in both of them
static void flipBitsOfAandB(int A, int B)
{
     
    // Clear the bits of A which
    // are set in both A and B
    A = A ^ (A & B);
     
    // Clear the bits of B which
    // are set in both A and B
    B = B ^ (A & B);
  
    // Print updated A and B
    System.out.print(A + " " + B);
}
    
// Driver Code
public static void main(String[] args)
{
    int A = 10, B = 20;
     
    flipBitsOfAandB(A, B);
}
}
 
// This code is contributed by sanjoy_62

Python3




# Python3 program to implement
# the above approach
 
# Function to flip bits of A and B
# which are set in both of them
def flipBitsOfAandB(A, B):
     
    # Clear the bits of A which
    # are set in both A and B
    A = A ^ (A & B)
 
    # Clear the bits of B which
    # are set in both A and B
    B = B ^ (A & B)
 
    # Print updated A and B
    print(A, B)
 
# Driver Code 
if __name__ == "__main__" :
     
    A = 10
    B = 20
     
    flipBitsOfAandB(A, B)
     
# This code is contributed by Virusbuddah_

C#




// C# program to implement
// the above approach
using System;
 
class GFG{
    
// Function to flip bits of A and B
// which are set in both of them
static void flipBitsOfAandB(int A, int B)
{
   
    // Clear the bits of A which
    // are set in both A and B
    A = A ^ (A & B);
     
    // Clear the bits of B which
    // are set in both A and B
    B = B ^ (A & B);
  
    // Print updated A and B
    Console.Write(A + " " + B);
}
    
// Driver Code
public static void Main(String[] args)
{
    int A = 10, B = 20;
     
    flipBitsOfAandB(A, B);
}
}
 
// This code is contributed by Amit Katiyar

Javascript




<script>
 
// javascript program to implement
// the above approach
 
    
// Function to flip bits of A and B
// which are set in both of them
function flipBitsOfAandB(A , B)
{
     
    // Clear the bits of A which
    // are set in both A and B
    A = A ^ (A & B);
     
    // Clear the bits of B which
    // are set in both A and B
    B = B ^ (A & B);
  
    // Print updated A and B
    document.write(A + " " + B);
}
    
// Driver Code
var A = 10, B = 20;
 
flipBitsOfAandB(A, B);
 
// This code contributed by shikhasingrajput
 
</script>
Output: 
10 20

 

Time Complexity: O(1) 
Auxiliary Space: O(1)

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